让面试官满意的排序算法（图文解析）

时间 2020-01-07

标签面试满意排序算法图文解析栏目快乐工作繁體版

原文原文链接

让面试官满意的排序算法（图文解析）

这种排序算法可以让面试官面露微笑
这种排序算法集各排序算法之大成
这种排序算法逻辑性十足
这种排序算法可以展现本身对Java底层的了解
这种排序算法出自Vladimir Yaroslavskiy、Jon Bentley和Josh Bloch三位大牛之手，它就是JDK的排序算法——java.util.DualPivotQuicksort（双支点快排）java

DualPivotQuicksort

先看一副逻辑图（若有错误请大牛在评论区指正）面试

插排指的是改进版插排—— 哨兵插排
快排指的是改进版快排——双支点快排算法

DualPivotQuickSort没有Object数组排序的逻辑，此逻辑在Arrays中，好像是归并+Tim排序segmentfault

图像应该很清楚：对于不一样的数据类型，Java有不一样的排序策略：数组

byte、short、char 他们的取值范围有限，使用计数排序占用的空间也不过256/65536个单位，只要排序的数量不是特别少（有一个计数排序阈值，低于这个阈值的话就没有不要用空间换时间了），都应使用计数排序
int、long、float、double 他们的取值范围很是的大，不适合使用计数排序
float和double 他们又有特殊状况：app
- NAN（not a number），NAN不等于任何数字，甚至不等于本身
- +0.0，-0.0，float和double没法精确表示十进制小数，咱们所看到的十进制小数其实都是取得近似值，于是会有+0.0（接近0的正浮点数）和-0.0（接近0的负浮点数)，在排序流程中统一按0来处理，于是最后要调整一下-0.0和+0.0的位置关系
Object

计数排序

计数排序是以空间换时间的排序算法，它时间复杂度O(n)，空间复杂度O(m)（m为排序数值可能取值的数量），只有在范围较小的时候才应该考虑计数排序less

（源码以short为例）工具

int[] count = new int[NUM_SHORT_VALUES]; //1 << 16 = 65536，即short的可取值数量

//计数，left和right为数组要排序的范围的左界和右界
//注意，直接把
for (int i = left - 1; ++i <= right;count[a[i] - Short.MIN_VALUE]++);

//排序
for (int i = NUM_SHORT_VALUES, k = right + 1; k > left; ) {
    while (count[--i] == 0);
    short value = (short) (i + Short.MIN_VALUE);
    int s = count[i];

    do {
        a[--k] = value;
    } while (--s > 0);
}

哨兵插排

当数组元素较少时，时间O(n^2^)和O(log~n~)其实相差无几，而插排的空间占用率要少于快排和归并排序，于是当数组元素较少时（<插排阈值），优先使用插排优化

哨兵插排是对插排的优化，原插排每次取一个值进行遍历插入，而哨兵插排则取两个，较大的一个（小端在前的排序）做为哨兵，当哨兵遍历到本身的位置时，另外一个值能够直接从哨兵当前位置开始遍历，而不用再重头遍历ui

只画了静态图，若是有好的绘制Gif的工具请在评论区告诉我哦

咱们来看一下源码：

if (leftmost) {
    //传统插排（无哨兵Sentinel）
    //遍历
    //循环向左比较（<左侧元素——换位）-直到大于左侧元素
    for (int i = left, j = i; i < right; j = ++i) {
        int ai = a[i + 1];
        while (ai < a[j]) {
            a[j + 1] = a[j];
            if (j-- == left) {
                break;
            }
        }
        a[j + 1] = ai;
    }
    
    //哨兵插排
} else {
    //若是一开始就是排好序的——直接返回
    do {
        if (left >= right) {
            return;
        }
    } while (a[++left] >= a[left - 1]);

    //以两个为单位遍历，大的元素充当哨兵，以减小小的元素循环向左比较的范围
    for (int k = left; ++left <= right; k = ++left) {
        int a1 = a[k], a2 = a[left];

        if (a1 < a2) {
            a2 = a1; a1 = a[left];
        }
        while (a1 < a[--k]) {
            a[k + 2] = a[k];
        }
        a[++k + 1] = a1;

        while (a2 < a[--k]) {
            a[k + 1] = a[k];
        }
        a[k + 1] = a2;
    }
    //确保最后一个元素被排序
    int last = a[right];

    while (last < a[--right]) {
        a[right + 1] = a[right];
    }
    a[right + 1] = last;
}
return;

双支点快排

重头戏：双支点快排！

快排虽然稳定性不如归并排序，可是它不用复制来复制去，省去了一段数组的空间，在数组元素较少的状况下稳定性影响也会降低（>插排阈值，<快排阈值），优先使用快排

双支点快排在原有的快排基础上，多加一个支点，左右共进，效率提高

看图：

第一步，取支点

注意：若是5个节点有相等的任两个节点，说明数据不够均匀，那就要使用单节点快排
快排

源码（int为例，这么长估计也没人看）

// Inexpensive approximation of length / 7 
// 快排阈值是286 其7分之一小于等于1/8+1/64+1
int seventh = (length >> 3) + (length >> 6) + 1;

// 获取分红7份的五个中间点
int e3 = (left + right) >>> 1; // The midpoint
int e2 = e3 - seventh;
int e1 = e2 - seventh;
int e4 = e3 + seventh;
int e5 = e4 + seventh;

// 保证中间点的元素从小到大排序
if (a[e2] < a[e1]) { 
    int t = a[e2]; a[e2] = a[e1]; a[e1] = t; }

if (a[e3] < a[e2]) { 
    int t = a[e3]; a[e3] = a[e2]; a[e2] = t;
    if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
}
if (a[e4] < a[e3]) { 
    int t = a[e4]; a[e4] = a[e3]; a[e3] = t;
    if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                    if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
                   }
}
if (a[e5] < a[e4]) { 
    int t = a[e5]; a[e5] = a[e4]; a[e4] = t;                    
    if (t < a[e3]) { a[e4] = a[e3]; a[e3] = t;
                    if (t < a[e2]) { a[e3] = a[e2]; a[e2] = t;
                                    if (t < a[e1]) { a[e2] = a[e1]; a[e1] = t; }
                                   }
                   }
}

// Pointers
int less  = left;  // The index of the first element of center part
int great = right; // The index before the first element of right part

//点彼此不相等——分三段快排，不然分两段
if (a[e1] != a[e2] && a[e2] != a[e3] && a[e3] != a[e4] && a[e4] != a[e5]) {
    /*
             * Use the second and fourth of the five sorted elements as pivots.
             * These values are inexpensive approximations of the first and
             * second terciles of the array. Note that pivot1 <= pivot2.
             */
    int pivot1 = a[e2];
    int pivot2 = a[e4];

    /*
             * The first and the last elements to be sorted are moved to the
             * locations formerly occupied by the pivots. When partitioning
             * is complete, the pivots are swapped back into their final
             * positions, and excluded from subsequent sorting.
             */
    a[e2] = a[left];
    a[e4] = a[right];

    while (a[++less] < pivot1);
    while (a[--great] > pivot2);

    /*
             * Partitioning:
             *
             *   left part           center part                   right part
             * +--------------------------------------------------------------+
             * |  < pivot1  |  pivot1 <= && <= pivot2  |    ?    |  > pivot2  |
             * +--------------------------------------------------------------+
             *               ^                          ^       ^
             *               |                          |       |
             *              less                        k     great
             */
    outer:
    for (int k = less - 1; ++k <= great; ) {
        int ak = a[k];
        if (ak < pivot1) { // Move a[k] to left part
            a[k] = a[less];
            /*
                     * Here and below we use "a[i] = b; i++;" instead
                     * of "a[i++] = b;" due to performance issue.
                     */
            a[less] = ak;
            ++less;
        } else if (ak > pivot2) { // Move a[k] to right part
            while (a[great] > pivot2) {
                if (great-- == k) {
                    break outer;
                }
            }
            if (a[great] < pivot1) { // a[great] <= pivot2
                a[k] = a[less];
                a[less] = a[great];
                ++less;
            } else { // pivot1 <= a[great] <= pivot2
                a[k] = a[great];
            }
            /*
                     * Here and below we use "a[i] = b; i--;" instead
                     * of "a[i--] = b;" due to performance issue.
                     */
            a[great] = ak;
            --great;
        }
    }

    // Swap pivots into their final positions
    a[left]  = a[less  - 1]; a[less  - 1] = pivot1;
    a[right] = a[great + 1]; a[great + 1] = pivot2;

    // Sort left and right parts recursively, excluding known pivots
    sort(a, left, less - 2, leftmost);
    sort(a, great + 2, right, false);

    /*
             * If center part is too large (comprises > 4/7 of the array),
             * swap internal pivot values to ends.
             */
    if (less < e1 && e5 < great) {
        /*
                 * Skip elements, which are equal to pivot values.
                 */
        while (a[less] == pivot1) {
            ++less;
        }

        while (a[great] == pivot2) {
            --great;
        }

        /*
                 * Partitioning:
                 *
                 *   left part         center part                  right part
                 * +----------------------------------------------------------+
                 * | == pivot1 |  pivot1 < && < pivot2  |    ?    | == pivot2 |
                 * +----------------------------------------------------------+
                 *              ^                        ^       ^
                 *              |                        |       |
                 *             less                      k     great
                 *
                 * Invariants:
                 *
                 *              all in (*,  less) == pivot1
                 *     pivot1 < all in [less,  k)  < pivot2
                 *              all in (great, *) == pivot2
                 *
                 * Pointer k is the first index of ?-part.
                 */
        outer:
        for (int k = less - 1; ++k <= great; ) {
            int ak = a[k];
            if (ak == pivot1) { // Move a[k] to left part
                a[k] = a[less];
                a[less] = ak;
                ++less;
            } else if (ak == pivot2) { // Move a[k] to right part
                while (a[great] == pivot2) {
                    if (great-- == k) {
                        break outer;
                    }
                }
                if (a[great] == pivot1) { // a[great] < pivot2
                    a[k] = a[less];
                    /*
                             * Even though a[great] equals to pivot1, the
                             * assignment a[less] = pivot1 may be incorrect,
                             * if a[great] and pivot1 are floating-point zeros
                             * of different signs. Therefore in float and
                             * double sorting methods we have to use more
                             * accurate assignment a[less] = a[great].
                             */
                    a[less] = pivot1;
                    ++less;
                } else { // pivot1 < a[great] < pivot2
                    a[k] = a[great];
                }
                a[great] = ak;
                --great;
            }
        }
    }

    // Sort center part recursively
    sort(a, less, great, false);

} else { // Partitioning with one pivot
    /*
             * Use the third of the five sorted elements as pivot.
             * This value is inexpensive approximation of the median.
             */
    int pivot = a[e3];

    /*
             * Partitioning degenerates to the traditional 3-way
             * (or "Dutch National Flag") schema:
             *
             *   left part    center part              right part
             * +-------------------------------------------------+
             * |  < pivot  |   == pivot   |     ?    |  > pivot  |
             * +-------------------------------------------------+
             *              ^              ^        ^
             *              |              |        |
             *             less            k      great
             *
             * Invariants:
             *
             *   all in (left, less)   < pivot
             *   all in [less, k)     == pivot
             *   all in (great, right) > pivot
             *
             * Pointer k is the first index of ?-part.
             */
    for (int k = less; k <= great; ++k) {
        if (a[k] == pivot) {
            continue;
        }
        int ak = a[k];
        if (ak < pivot) { // Move a[k] to left part
            a[k] = a[less];
            a[less] = ak;
            ++less;
        } else { // a[k] > pivot - Move a[k] to right part
            while (a[great] > pivot) {
                --great;
            }
            if (a[great] < pivot) { // a[great] <= pivot
                a[k] = a[less];
                a[less] = a[great];
                ++less;
            } else { // a[great] == pivot
                /*
                         * Even though a[great] equals to pivot, the
                         * assignment a[k] = pivot may be incorrect,
                         * if a[great] and pivot are floating-point
                         * zeros of different signs. Therefore in float
                         * and double sorting methods we have to use
                         * more accurate assignment a[k] = a[great].
                         */
                a[k] = pivot;
            }
            a[great] = ak;
            --great;
        }
    }

    /*
             * Sort left and right parts recursively.
             * All elements from center part are equal
             * and, therefore, already sorted.
             */
    sort(a, left, less - 1, leftmost);
    sort(a, great + 1, right, false);
}

归并排序

你不会觉得元素多（>快排阈值）就必定要用归并了吧？

错！元素多时确实对算法的稳定性有要求，但是若是这些元素可以稳定快排呢？

开发JDK的大牛显然考虑了这一点：他们在归并排序以前对元素进行了是否能稳定快排的判断：

若是数组自己几乎已经排好了（能够看出几段有序数组的拼接），那还排什么，理一理返回就好了
若是出现连续33个相等元素——使用快排（实话说，我没弄明白为何，有无大牛给我指点迷津？）

//判断结构是否适合归并排序
int[] run = new int[MAX_RUN_COUNT + 1];
int count = 0; run[0] = left;

// Check if the array is nearly sorted
for (int k = left; k < right; run[count] = k) {
    if (a[k] < a[k + 1]) { // ascending
        while (++k <= right && a[k - 1] <= a[k]);
    } else if (a[k] > a[k + 1]) { // descending
        while (++k <= right && a[k - 1] >= a[k]);
        for (int lo = run[count] - 1, hi = k; ++lo < --hi; ) {
            int t = a[lo]; a[lo] = a[hi]; a[hi] = t;
        }
    } else { 
        //连续MAX_RUN_LENGTH（33）个相等元素，使用快排
        for (int m = MAX_RUN_LENGTH; ++k <= right && a[k - 1] == a[k]; ) {
            if (--m == 0) {
                sort(a, left, right, true);
                return;
            }
        }
    }

    //count达到MAX_RUN_LENGTH，使用快排
    if (++count == MAX_RUN_COUNT) {
        sort(a, left, right, true);
        return;
    }
}

// Check special cases
// Implementation note: variable "right" is increased by 1.
if (run[count] == right++) { // The last run contains one element
    run[++count] = right;
} else if (count == 1) { // The array is already sorted
    return;
}

归并排序源码

byte odd = 0;
for (int n = 1; (n <<= 1) < count; odd ^= 1);

// Use or create temporary array b for merging
int[] b;                 // temp array; alternates with a
int ao, bo;              // array offsets from 'left'
int blen = right - left; // space needed for b
if (work == null || workLen < blen || workBase + blen > work.length) {
    work = new int[blen];
    workBase = 0;
}
if (odd == 0) {
    System.arraycopy(a, left, work, workBase, blen);
    b = a;
    bo = 0;
    a = work;
    ao = workBase - left;
} else {
    b = work;
    ao = 0;
    bo = workBase - left;
}

// Merging
for (int last; count > 1; count = last) {
    for (int k = (last = 0) + 2; k <= count; k += 2) {
        int hi = run[k], mi = run[k - 1];
        for (int i = run[k - 2], p = i, q = mi; i < hi; ++i) {
            if (q >= hi || p < mi && a[p + ao] <= a[q + ao]) {
                b[i + bo] = a[p++ + ao];
            } else {
                b[i + bo] = a[q++ + ao];
            }
        }
        run[++last] = hi;
    }
    if ((count & 1) != 0) {
        for (int i = right, lo = run[count - 1]; --i >= lo;
             b[i + bo] = a[i + ao]
            );
        run[++last] = right;
    }
    int[] t = a; a = b; b = t;
    int o = ao; ao = bo; bo = o;
}

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